= Solution
For the <adjacent reflected minima of an air-film-air layer> in <thin-film interference>, reflection at the front surface acquires a phase reversal, while reflection at the rear does not. Near normal incidence the reflected minima therefore obey $2nt=m\lambda$ for integral m. A single intermediate maximum means that the two stated minima are consecutive orders:
$$
\frac{2nt}{500\,\mathrm{nm}}-\frac{2nt}{600\,\mathrm{nm}}=1.
$$
Thus $2nt=3000\,\mathrm{nm}$, with orders6 and5, and
$$
\boxed{t=\frac{3000}{2(1.3)}\,\mathrm{nm}=1.15\,\mu\mathrm m.}
$$
The intermediate reflected maximum is at $\lambda=3000/5.5\simeq545\,\mathrm{nm}$. This assumes negligible dispersion of the supplied refractive index and an air environment on both sides. <University Physics Volume 3 by Samuel J. Ling and others>[§3.4, reflection phase reversals and thin-film interference]{ref}
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